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A130477 T(n,k) is the number of permutations of [n] with maximum descent k, T(n,k) for n >= 0 and 0 <= k <= n, triangle read by rows.

%I A130477 #36 Sep 17 2018 06:54:20
%S A130477 1,1,1,1,2,3,1,3,8,12,1,4,15,40,60,1,5,24,90,240,360,1,6,35,168,630,
%T A130477 1680,2520,1,7,48,280,1344,5040,13440,20160,1,8,63,432,2520,12096,
%U A130477 45360,120960,181440,1,9,80,630,4320,25200,120960,453600,1209600,1814400,1,10,99,880,6930,47520,277200,1330560,4989600,13305600,19958400
%N A130477 T(n,k) is the number of permutations of [n] with maximum descent k, T(n,k) for n >= 0 and 0 <= k <= n, triangle read by rows.
%C A130477 Previous name was: Triangle generated from finite differences of A130461.
%C A130477 Decomposition of the permutations.
%C A130477 From _Liam Solus_, Aug 23 2018: (Start)
%C A130477 For k>0, T(n,k) equals the number of permutations p of [n] such that the largest index i for which p(i)>p(i+1) is k; i.e., T(n,k) is the number of permutations of [n] with maximum descent being k. See Lemma 3.4 of the paper by L. Solus below.
%C A130477 When T(n,k) is taken as the weight of coordinate x_k for k = 0,...,n-1 in an (n-1)-dimensional weighted projective space, the result is the toric variety defined by an n-dimensional simplex whose Ehrhart h^*-polynomial is the n-th Eulerian polynomial. See Theorem 3.5 of the paper by L. Solus below.
%C A130477 (End)
%H A130477 L. Solus, <a href="https://arxiv.org/abs/1706.00480">Simplices for numeral systems</a>, Transactions of the American Mathematical Society. DOI: https://doi.org/10.1090/tran/7424 (2017).
%F A130477 Each term in n-th row divides n!.
%F A130477 Given triangle A130461 and deleting the left border (1,1,1,...) take finite differences by columns and reorient into rows.
%F A130477 T(n,k) = (n-k+1+0^k)*((n+1)!/(n-k+2)!) - _Olivier Gérard_, Aug 04 2012
%e A130477 First few rows of the triangle A130461 = (1; 1, 1; 1, 1, 1; 1, 1, 2, 1; 1, 1, 2, 3, 1; 1, 1, 2, 6, 4, 1;...). Deleting the left border and taking finite differences at the top of each remaining column, we get the first few rows of this triangle:
%e A130477 1;
%e A130477 1, 1;
%e A130477 1, 2,  3;
%e A130477 1, 3,  8,  12;
%e A130477 1, 4, 15,  40,  60;
%e A130477 1, 5, 24,  90, 240,  360;
%e A130477 1, 6, 35, 168, 630, 1680, 2520;
%e A130477 ...
%p A130477 T := (n,k) -> (n-k+1+0^k)*((n+1)!/(n-k+2)!):
%p A130477 seq(seq(T(n,k),k=0..n),n=0..10); # _Peter Luschny_, Sep 17 2018
%t A130477 Flatten[Table[Table[(n - k + 1 + 0^k)*(n + 1)!/(n - k + 2)!, {k,0,n}], {n, 0, 10}], 1] (* _Olivier Gérard_, Aug 04 2012 *)
%o A130477 (PARI) {T(n, k) = if( n<1 || k>n, 0, k==1, 1, n! * (n+1-k) / (n+2-k)!)}; /* _Michael Somos_, Jun 25 2017 */
%Y A130477 The triangle with each row reversed is A092582.
%Y A130477 Cf. A000142 (row sums), A001710 (main diagonal), A008292.
%Y A130477 Cf. A130460, A130461, A130476, A130478.
%K A130477 nonn,tabl
%O A130477 0,5
%A A130477 _Gary W. Adamson_, May 28 2007
%E A130477 New name using a comment by _Liam Solus_, _Peter Luschny_, Sep 17 2018